The vapour pressure of pure liquids `A` and `B` is 450 and `700mm Hg`, respectively, at `350K. ` Find out the composition of the liquid mixture if the total vapour pressure is `600mm Hg`. Also find the composition of the vapour phase.

Explanation

Let mole fraction of liquid A in solution = `x_(A)`
Mole fraction of liquid B is solution = `x_(B) = (1 - x_(A))`
`p = p_(A) + p_(B) = x_(A)p_(A)^(@) + x_(B)P_(B)^(@)`
`x_(A)P_(A)^(@) + (1-x_(A)) p_(B)^(@) = x_(A)p_(A)^(@) + p_(B)^(@) - x_(A)p_(B)^(@)`
`therefore " "600 =x_(A) xx 450 + 700 - x_(A) xx 700`
`250 x_(A) = 100`
`therefore " "600 = x_(A) xx 450 + 700 - x_(A) xx 700`
`250 x_(A) = 100`
`therefore " "x_(A) = (100)/(250) = 0.4`
`" "x_(B) = 1 - 0.4 = 0.6`
Composition in vapour phase.
`p_(A) = x_(A)p_(A)^(@) = 0.4 xx 450 = 180 mm Hg`
`p_(B) = x_(B)p_(B)^(@) = 0.6 xx 700 = 420 mm Hg`
Total pressure = 1800 + 420 = 600 mm
Mole fraction of A in vapour phase , `y_(A) = (180)/(600) = 0.3`
Mole fraction of B in vapour phase . `y_(B) = (420)/(600) = 0.7`